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Chemistry

Vapor pressure (Clausius–Clapeyron) calculator

Finds vapor pressure at a new temperature from a known point and enthalpy of vaporization.

Published 25 September 2026

What this calculator does

The Clausius-Clapeyron equation predicts vapour pressure at one temperature from a measurement at another, given the enthalpy of vaporisation. From water boiling at 1 atm at 373.15 K, it predicts 0.0369 atm at 298.15 K.

That prediction is about 28 mmHg against a true value near 23.8, an overestimate of roughly 18%. The error comes from assuming the enthalpy of vaporisation is constant, when in reality it falls as temperature rises. The equation is reliable over modest temperature ranges and degrades over large ones, which a 75 K extrapolation certainly is.

The formula

Formulaln(P₂/P₁) = −(ΔHvap/R) × (1/T₂ − 1/T₁), R = 8.314 J/(mol·K)

The log of the pressure ratio equals minus the enthalpy of vaporisation over R, multiplied by the difference in reciprocal temperatures. R is 8.314 J/(mol·K), so the enthalpy must be in joules per mole. Both temperatures must be absolute.

TermMeaning
ΔHvapEnthalpy of vaporisation, the energy needed to convert a mole of liquid to vapour.
Vapour pressureThe pressure of vapour in equilibrium with its liquid at a given temperature.
Normal boiling pointThe temperature at which vapour pressure reaches 1 atm.
Constant ΔHvap assumptionThe main approximation, which weakens over wide temperature ranges.

The inputs explained

FieldWhat to enter
Known vapor pressure P₁ (atm)Known vapour pressure at the reference temperature.
Known temperature T₁ (K)Reference temperature in kelvin.
Enthalpy of vaporization ΔHvap (J/mol)Enthalpy of vaporisation in J/mol. Water is about 40,700.
New temperature T₂ (K)Temperature at which you want the vapour pressure, in kelvin.

When to use it

Estimating vapour pressure

Predicting a value at a temperature where no measurement exists, from a known boiling point.

Finding an enthalpy of vaporisation

Two vapour pressure measurements at different temperatures give ΔHvap by rearrangement.

Predicting a boiling point at altitude

Setting the target pressure to the local atmospheric pressure gives the boiling temperature there.

Worked examples

Every figure in the tables below is produced by this page’s own calculator at build time, so the numbers and the tool always agree. Select any row to load that scenario.

How does vapour pressure fall with temperature?

Predicted vapour pressure at a range of temperatures.

Water, 1 atm at 373.15 K, ΔHvap 40,700 J/mol
TemperatureVapor pressure at T₂Reference pressure at T₁Note
273.15 K0.00820626 atm1.000 atmAssumes ΔHvap is constant over the temperature range and vapor behaves ideally.
298.15 K0.0368794 atm1.000 atmAssumes ΔHvap is constant over the temperature range and vapor behaves ideally.
323.15 K0.13135348 atm1.000 atmAssumes ΔHvap is constant over the temperature range and vapor behaves ideally.
373.15 K1 atm1.000 atmAssumes ΔHvap is constant over the temperature range and vapor behaves ideally.
The last row returns exactly 1 atm, as it must, since it asks for the pressure at the reference temperature itself. That is a useful sanity check. The 298.15 K prediction of 0.0369 atm is about 28 mmHg against a true value near 23.8, an overestimate from assuming constant ΔHvap over a 75 K range.

Questions

Why is the prediction not exact?

Because the equation assumes the enthalpy of vaporisation is constant, when it actually decreases as temperature rises. Over a small range the error is negligible; over the 75 K used in the table above it reaches about 18%. Shorter extrapolations are much more reliable.

Must the temperatures be in kelvin?

Yes. The equation uses reciprocal absolute temperature, so Celsius values give a meaningless result. This is the most common error in applying it, and the answer it produces is wrong without being obviously so.

Can I use this to find the enthalpy of vaporisation?

Yes, by rearranging. Two vapour pressure measurements at two temperatures determine ΔHvap, in the same way the two-point Arrhenius equation determines an activation energy. The mathematical structure is identical.

How does this explain cooking at altitude?

Atmospheric pressure falls with altitude, so water boils when its vapour pressure reaches that lower value, which happens at a lower temperature. At 3,000 m water boils near 90 °C, which is why food takes longer to cook: it is simply not as hot.

For vapour pressure lowering by a solute, see the Raoult’s law calculator. For boiling point at altitude, see the boiling point at altitude calculator.